Question 2 of 8: Open-circuited stub matching a parallel RC load
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, May 2015 —
07-Elec-A7 Electromagnetics. Three hours, closed book; one of two approved
calculators (Casio or Sharp) permitted. Eight questions, all of equal value; the
paper’s note 3 states that any five questions constitute a complete paper
and only the first five presented will be marked. Because this set is a study
resource, all eight questions are solved here. The permitted aids
printed on the cover are used throughout:
$\varepsilon_0 = 8.85\times10^{-12}\ \text{F/m}$ and
$\mu_0 = 4\pi\times10^{-7}\ \text{H/m}$, giving
$\eta_0 = \sqrt{\mu_0/\varepsilon_0} = 376.6\ \Omega$ and
$c = 1/\sqrt{\mu_0\varepsilon_0} \approx 3\times10^{8}\ \text{m/s}$.
Reference texts. D. M. Pozar, Microwave Engineering, 4th ed.
(transmission-line transients, matching, waveguides); M. N. O. Sadiku,
Elements of Electromagnetics, 7th ed. (magnetostatics, Faraday’s law,
plane waves); W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed.
(polarisation, boundary conditions); C. A. Balanis, Antenna Theory: Analysis
and Design, 4th ed. (short elements, image theory); F. T. Ulaby and U. Ravaioli,
Fundamentals of Applied Electromagnetics, 8th ed. (waves and power density).
Given. A 300 MHz, 50 $\Omega$ generator feeds a parallel RC load through 50 $\Omega$ line, and an open-circuited stub of identical line is bridged across that load.
Given data
Quantity
Symbol
Value
Frequency
$f$
$300\ \text{MHz}$
Characteristic impedance (line and stub)
$Z_0$
$50\ \Omega$
Phase velocity
$v_p$
$3\times10^{8}\ \text{m/s}$
Load resistance
$R$
$50\ \Omega$
Load capacitance
$C$
$10.7\ \text{pF}$
Generator internal impedance
$R_g$
$50\ \Omega$
Find. The shortest length $l$ of the open-circuited stub that makes the admittance at the load plane equal to $Y_0 = 1/Z_0$, so that the line, load and generator are all matched.
Figure 2.1 — the open-circuited shunt stub bridged across the parallel RC load at the end of the 50 Ω feeder.
Approach. Because the stub is in parallel with the load, work entirely in admittances: normalise the load admittance to $Y_0$, note that its conductance is already unity, and choose the stub length whose purely imaginary input admittance cancels the leftover susceptance.
Wavelength on the line. The stub and feeder share $v_p$, so $$\lambda = \frac{v_p}{f} = \frac{3\times10^{8}}{300\times10^{6}} = 1.000\ \text{m},\qquad \beta = \frac{2\pi}{\lambda} = 6.283\ \text{rad/m}$$
Normalised load admittance. A resistor and capacitor in parallel add as admittances directly: $$Y_L = \frac{1}{R} + j\omega C, \qquad y_L = \frac{Y_L}{Y_0} = \frac{Z_0}{R} + j\,Z_0\,\omega C$$ With $\omega = 2\pi(300\times10^{6}) = 1.885\times10^{9}\ \text{rad/s}$, $$y_L = 1 + j\,(50)(1.885\times10^{9})(10.7\times10^{-12}) = 1 + j1.0084$$
Read what the stub has to do. The real part of $y_L$ is already exactly 1, because the load resistor equals $Z_0$. Matching therefore needs no line transformation at all — the stub only has to supply $-j1.0084$ at the same plane: $$y_{\text{stub}} = -j\,b_L, \qquad b_L = Z_0\omega C = 1.0084$$
Input admittance of an open stub. An open-circuited length $l$ of lossless line presents $$y_{\text{stub}} = j\tan\beta l \;\Longrightarrow\; \tan\beta l = -1.0084$$ A negative tangent means the stub must be longer than a quarter wavelength, so that its open end has already transformed into an inductive (negative-susceptance) presentation.
Solve for the length. Taking the first root beyond $\pi/2$, $$\beta l = \pi - \arctan(1.0084) = \pi - 0.7918 = 2.3520\ \text{rad} = 134.76^\circ$$$$l = \frac{\beta l}{\beta} = \frac{2.3520}{6.283} = \boxed{0.374\ \text{m} \;=\; 0.374\lambda \;=\; 37.4\ \text{cm}}$$
Confirm the match. Substituting back, $y_{\text{total}} = 1 + j1.0084 - j1.0084 = 1$, so $Z_{\text{in}} = Z_0 = 50\ \Omega$ at the load plane. A matched load makes the feeder reflectionless at every plane, so the generator (also 50 $\Omega$) sees 50 $\Omega$ no matter how long the feeder is, and delivers its full available power.
Because the tangent is periodic, every length $l + n\lambda/2 = 0.374,\ 0.874,\ 1.374\ \text{m},\ \ldots$ matches equally well; the shortest is quoted as the design value. Note also that the answer depends only on the product $Z_0\omega C$, which is why a stub of the same characteristic impedance is specified — a stub of different $Z_0$ would need a different length.
